Question

let zi(i=1,2,3,4) represent the vertices of a square all of which lie on the sides of the triangle with vertices (0,0),(2,1),(3,0) if z1 and z2 are purely real then area of triangle formed by z3 and z4 and origin is m/n (where m and n are in their lowest form). find the value of m+n

Answer 1

Answer: 41

Step-by-step explanation: Let z₁ be at a distance 'a' units from O.

Then coordinates of z₁ are (a,0).

Let 's' be the side of the square.

Then coordinates of z₂ are (a+s,0).

Coordinates of z₃ are (a+s,s) and

Coordinates of z₄ are (a,s).

Also, z₄ lies on OB

=> s/a = 1/2 =>2s =a

z₃ lies on AB

=> s/s+a-3 = -1

=>s+a-3 = -s

=>s =3/4 and a = 3/2

So, the vertices of z₃ are (3/2, 3/4)

vertices of z₄ are (9/4,3/4)

Area of triangle Oz₃z₄ = 1/2*s*s =9/32=m/n(given)

Hence, m=9, n=32

So, m+n =41.